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Math Help - Chain Rule?

  1. #1
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    Chain Rule?

    Given that x = e^t, find \frac{dy}{dx} in terms of \frac{dy}{dt} and show that

    \frac{d^2y}{dx^2} = e^{-2}\left(\frac{d^2y}{dt^2} -\frac{dy}{dt}\right).



    EDIT: Sorry, It is a typo, it should be \frac{d^2y}{dx^2} = e^{-2t}\left(\frac{d^2y}{dt^2} -\frac{dy}{dt}\right).
    Last edited by Simplicity; June 15th 2008 at 02:49 AM.
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  2. #2
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    I remember doing this question!
    *dives into the past papers*




    I also remember doing it wrong and teacher correcting me

    I can do the first part though...

    Using the chain rule, we know:
    \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=  \frac{dy}{dt} . \frac{dt}{dx}

    x=e^t \\ \Longrightarrow \\ \frac{dx}{dt}=e^t \\ \Longrightarrow \\ \frac{dt}{dx}=\frac{1}{e^t}

    and if you substitute that in, you'll get this:

    \frac{dy}{dx}=\frac{1}{e^t}.\frac{dy}{dt}
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  3. #3
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    Quote Originally Posted by Silver View Post


    I remember doing this question!
    *dives into the past papers*




    I also remember doing it wrong and teacher correcting me

    I can do the first part though...

    Using the chain rule, we know:
    \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=  \frac{dy}{dt} . \frac{dt}{dx}

    x=e^t \\ \Longrightarrow \\ \frac{dx}{dt}=e^t \\ \Longrightarrow \\ \frac{dt}{dx}=\frac{1}{e^t}

    and if you substitute that in, you'll get this:

    \frac{dy}{dx}=\frac{1}{e^t}.\frac{dy}{dt}
    Yes, I got that too. But it's the next stage that is tricky. How do I differentiate that:

    \frac{dy}{dx}=\frac{1}{e^t}.\frac{dy}{dt}

    ?
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  4. #4
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    Quote Originally Posted by Air View Post
    Yes, I got that too. But it's the next stage that is tricky. How do I differentiate that:

    \frac{dy}{dx}=\frac{1}{e^t}.\frac{dy}{dt}

    ?
    \frac{d^2 y}{dx^2} = \frac{d}{dx} \left( e^{-t} \, \frac{dy}{dt} \right)


    = e^{-t} \frac{d}{dx} \left( \frac{dy}{dt} \right) + \frac{dy}{dt} \frac{d}{dx} \left( e^{-t} \right)

    using the product rule


    = e^{-t} \frac{d}{dt} \left( \frac{dy}{dt} \right) \, \frac{dt}{dx}+ \frac{dy}{dt} \, \frac{d}{dt} \left( e^{-t} \right) \frac{dt}{dx}

    using the chain rule


    = e^{-t} \, \frac{d^2y}{dt^2} \, e^{-t} + \frac{dy}{dt} \, \left(-e^{-t} \right) \, e^{-t}


    = e^{-2t} \, \frac{d^2y}{dt^2} - \frac{dy}{dt} \, e^{-2t} .....
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  5. #5
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    Hi,

    I'm not really sure because I can't get to your equation...
    And sorry for the "d", it takes too long time to straighten them...

    {\color{red}\frac{dy}{dx}}=e^{-t} \frac{dy}{dt}


    You're looking for \frac{d^2y}{dx^2}

    -----------------------

    \begin{aligned} \frac{d^2y}{dx^2} &=\frac{d}{dx} \left({\color{red}\frac{dy}{dx}}\right) \\ \\<br />
&=\frac{d}{\color{green}dx} \left(e^{-t} \frac{dy}{dt}\right) \end{aligned}

    Now, x=e^t \implies {\color{green}dx}=e^t dt (we have to do it because the variable is t in the brackets, not x).

    Replacing :

    \begin{aligned} \frac{d^2y}{dx^2} &=\frac{d}{\color{green}dx} \left(e^{-t} \frac{dy}{dt}\right) \\<br />
&=\frac{d}{e^t dt} \left(e^{-t} \frac{dy}{dt}\right) \\<br />
&=e^{-t} \frac{d}{dt} \left(e^{-t} \frac{dy}{dt} \right)<br />
\end{aligned}

    Using the product rule :

    \begin{aligned} \frac{d^2y}{dx^2} &=e^{-t} \left(-e^{-t} \frac{dy}{dt}+e^{-t} \frac{d^2y}{dt^2} \right) \\<br />
&=e^{-2t} \left(-\frac{dy}{dt}+\frac{d^2y}{dt^2}\right) \end{aligned}




    Edit : oh I see, you made a typo, Air

    Re-edit : too fast for me Mr F
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  6. #6
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    I think You have written the second part wrong. It should be:
     <br />
\frac{d^2y}{dx^2} = e^{-2t}\left(\frac{d^2y}{dt^2} -\frac{dy}{dt}\right)<br />
    Solution:
     <br />
\frac{dy}{dx}=\frac{1}{x}\frac{dy}{dt}<br />
     <br />
\frac{d^2y}{dx^2}=\frac{d}{dx}\left(\frac{dy}{dx}\  right)=\frac{d}{dx}\left(\frac{1}{x}\frac{dy}{dt}\  right)=\frac{1}{x}\frac{d}{dx}\frac{dy}{dt}-\frac{1}{x^2}\frac{dy}{dt}=\frac{1}{x}\frac{d}{dt}  \frac{dy}{dx}-\frac{1}{x^2}\frac{dy}{dt}=\frac{1}{x^2}\frac{d^2y  }{dt^2}-\frac{1}{x^2}\frac{dy}{dt}<br />
     <br />
=e^{-2t}\left(\frac{d^2y}{dt^2} -\frac{dy}{dt}\right)<br />

    Regards,
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